Laurent polynomials and Eulerian numbers

0.1 MB·English

by  Daniel Erman

LAURENT POLYNOMIALS AND EULERIAN NUMBERS
DANIELERMAN,GREGORYG.SMITH,ANDANTHONYVÁRILLY-ALVARADO
0 ABSTRACT. DuistermaatandvanderKallenshowthatthereis nonontrivialcomplexLaurent
1
polynomialallofwhosepowershaveazeroconstantterm.Inspiredbythis,Sturmfelsposestwo
0
questions:DotheconstanttermsofagenericLaurentpolynomialformaregularsequence?Ifso,
2
thenwhatisthedegreeoftheassociatedzero-dimensionalideal? Inthisnote,weprovethatthe
n
Euleriannumbersprovidetheanswertothesecondquestion. Theproofinvolvesreinterpreting
a
J theproblemintermsoftoricgeometry.

4
2

]
O 1. MOTIVATION AND STATEMENT OF THEOREM
C
In [DK], J.J. Duistermaat and W. van der Kallen establish that, for any Laurent polynomial
h. f ∈C[z,z−1]thatisneitherapolynomialinznorz−1,thereexistsapositivepowerof f thathas
t
a a nonzero constant term. Motivated by this result, Sturmfels [Stu, §2.5] asks for an effective
m
version: Can weenumeratetheLaurent polynomialsthathavethelongestpossiblesequenceof
[
powerswithzero constantterms?

2 By rephrasing this question in the language of commutativealgebra, Sturmfels also offers a
v
two-stepapproach foransweringit. Specifically,considertheLaurentpolynomial
9
0
f(z):=z−m+x z−m+1+···+x zn−1+zn (♠)
6 −m+1 n−1
2
and, for any positive integer i, let JfiK denote the constant coefficient of the i-th power of f.
.
8
First,Problem2.11in[Stu,§2.5],togetherwithcomputationalevidence,suggeststhefollowing:
0
9
Conjecture 1. ThecoefficientsJf1K,Jf2K,...,Jfm+nKgeneratetheunitidealinthepolynomial
0
: ringC[x ,...,x ].
v −m+1 n−1
i
X Second, assuming this conjecture, Exercise 13 in [Stu, §2.6] asks for the degree of the ideal
r I := Jf1K,Jf2K,...,Jfm+n−1K . The zeros of I would be the Laurent polynomials of the
a m,n m,n
form (♠)that havethelongestpossiblesequenceofpowerswithvanishingconstantterms.
(cid:10) (cid:11)
The goal of this article is to complete the second part. Theorem 2 provides the unexpected
n
and attractivelysimpleanswer. Following[GKP, §6.2], the Eulerian number is the number
k
ofpermutationsof{1,...,n}withexactly k ascents.
(cid:10) (cid:11)
Theorem 2. IfConjecture1 holds,thenthedegreeof theidealI is m+n−1 .
m,n m−1
This result is equivalentto saying that thedimension of C[x−m+1,...,(cid:10)xn−1]/I(cid:11)m,n, as a C-vector
space, is m+n−1 .
m−1
Notably,Theorem2givesanewinterpretationfortheEuleriannumbers: m+n−1 enumerates
(cid:10) (cid:11) m−1
certain Laurentpolynomials. EvenwithoutConjecture1,weshowthattheseEuleriannumbers
(cid:10) (cid:11)
2000MathematicsSubjectClassification. 05A10,14N10,14M25.
1
2 D.ERMAN,G.G.SMITH,ANDA.VÁRILLY-ALVARADO
count the solutions to certain systems of polynomial equations; see Proposition 4. Despite su-
perficial similaritiesbetween ourwork and otherappearances ofEuleriannumbersin algebraic
geometry(e.g. [BS,Bre,BW,Hir,Sta,Ste]), weknowofno substantiveconnection.
Our proof of Theorem 2, given in §2, recasts the problem in terms of toric geometry — we
construethedegreeofI asanintersectionnumberonatoriccompactificationofthespaceof
m,n
Laurentpolynomialsoftheform(♠). Buildingonthisidea,§3providesarecursiveformulafor
thedegreeofidealssimilartoI thatarisefromsparseLaurentpolynomials. Asaby-product,
m,n
we give a geometric explanation for a formula expressing m+n−1 as a sum of nonnegative
m−1
integers;see(♥). Welistseveralquestionsarisingfromourwork in§4.
(cid:10) (cid:11)
ACKNOWLEDGEMENTS. We thank David Eisenbud, Alexander Postnikov, Bernd Sturmfels,
and Mauricio Velasco for useful conversations. The computer software Macaulay 2 [M2] was
indispensable in discovering both the statement and proof of the theorem. The Mathematical
Sciences Research Institute (MSRI), in Berkeley, provided congenial surroundings in which
much of this work was done. Erman was partially supported by a NDSEG fellowship, Smith
was partially supported by NSERC, and Várilly-Alvarado was partially supported by a Marie
CurieResearch TrainingNetworkwithintheSixthEuropeanCommunityFrameworkProgram.
2. TORIC REINTERPRETATION
This section proves Theorem 2 by reinterpreting the degree of I as an intersection number
m,n
on a projectivevariety X(m,n). Subsection §2.1 introduces a homogenization of the ideal I ,
m,n
§2.2describes thetoricvarietyX(m,n),and §2.3computestherequired intersectionnumber.
2.1. HOMOGENIZATION. Forpositiveintegersm andn, considertheLaurent polynomial
f˜:=x z−m+x z−m+1+···+x zn−1+x zn,
−m −m+1 n−1 n
and,foranypositiveintegeri,letJf˜iKdenotetheconstantcoefficientofthei-thpowerof f˜. Let
SbethepolynomialringC[x ,...,x ]andletJ betheS-ideal Jf˜1K,Jf˜2K,...,Jf˜m+n−1K . The
−m n
C-valuedpointsofV(J)⊂Am+n+1arepreciselytheLaurentpolynomialsforwhichtheconstant
(cid:10) (cid:11)
termofthefirst m+n−1 powersvanishes. SinceJ iscontainedinthereduced monomialideal
B := hx ,...,x i∩hx ,x ,...,x i, the C-valued points of V(J) not contained in V(B) give
−m −1 0 1 n
risetoLaurent polynomialsthat areneitherpolynomialsin z norz−1.
TounderstandtheidealJ moreexplicitly,letw:=[−m ··· n]t∈Zm+n+1. Ifu∈Nm+n+1,then
themultinomialtheorem[GKP,p. 168]impliesthat
Jf˜iK= (cid:229) i xu = (cid:229) i xu1 xu2 ···xum+n+1.
u u ,...,u −m −m+1 n
|u|=i (cid:18) (cid:19) |u|=i (cid:18) 1 m+n+1(cid:19)
w·u=0 w·u=0
Hence,forallpositiveintegersi,thepolynomialJf˜iKishomogeneousofdegree i withrespect
0
totheZ2-gradingofSinducedbysettingdeg(x ):= 1 ∈Z2 forall−m6 j6n. Inparticular,
j j (cid:2) (cid:3)
the ideal J is invariant under the automorphism of S determined by the map f˜(z) 7→ l f˜(x z)
(cid:2) (cid:3)
where l ,x ∈C∗. Moreover, if x and x are both nonzero, then there exist scalars l , x ∈C∗
−m n
such thattheimageof f˜underthis(C∗)2-actionhas theform(♠).
LAURENTPOLYNOMIALSANDEULERIANNUMBERS 3
2.2. TORIC VARIETY. When m+n >2, let X(m,n) be the toric variety with total coordinate
ring S (a.k.a. the Cox ring) and irrelevant ideal B; see [Cox, §2]. The variety X(m,n)provides
a toric compactification for the space of all Laurent polynomials of the form (♠). When no
confusion is likely, we simply write X in place of X(m,n). Proposition 2.4 in [Cox] shows
thathomogeneousS-ideals(uptoB-torsion)correspondtoclosedsubschemesofX. Hence,the
idealJ determinesaclosedsubschemeV (J)ofX. Ifx x isanonzerodivisoronV (J),then
X −m n X
§2.1showsthatthedegreeoftheidealI equalsthedegreeofV (J). WeproveTheorem2by
m,n X
computingthelatterdegree.
More concretely, X is the toric variety associated to the following strongly convex rational
polyhedral fan S ; see [Fu1, §1.4]. The lattice of one-parameter subgroups is N =Zm+n−1 and
therays (i.e. one-dimensionalcones)in thefan S are generated by thecolumnsofthematrix:
1 −2 1 0 ··· 0
2 −3 0 1 ··· 0
 .. .. .. .. .. ... (♣)
. . . . . .
 
m+n−1 −m−n 0 0 ··· 1
 
With the column ordering, we label the rays in S by r ,...,r. For integers 1 6 i 6 m and
−m n
06 j 6n, let s be the cone in Rm+n−1 =N⊗ R spanned by all the rays except r and r .
i,j Z −i j
Thefan S isdefined bytakingtheses asthemaximalcones. By construction,X isasingular
i,j
simplicialprojectivetoricvarietyofdimensionm+n−1.
2.3. INTERSECTION THEORY. Since X is a simplicial toric variety, its rational Chow ring
A∗(X) has an explicit presentation; see [Fu1, §5.2]. Specifically, if D is the torus-invariant
Q j
Weil divisorassociatedto theray r forall −m6 j6n,then wehave
j
Q[D ,...,D ]
A∗(X) = −m n
Q
M+L
where the monomial ideal M :=hD D ···D ,D D ···D D i is the Alexander dual
−m −m+1 −1 0 1 n−1 n
of B, and the linear ideal L:=hiD −(i+1)D +D :16i6m+n−1i encodes
−m −m+1 −m+i+1
therowsofthematrix(♣).
Choosing a shelling for the fan S yields a distinguished basis for A∗(X) ; again see [Fu1,
Q
§5.2]. With this in mind, we order the maximal cones of S by s > s if i+ j > k+ℓ or
i,j k,ℓ
i+ j = k+ℓ and j > ℓ. Let t be the subcone of s obtained by intersecting the maximal
i,j i,j
cone s with all cones s satisfying s >s and dims ∩s =m+n−2. We obtain a
i,j k,ℓ k,ℓ i,j i,j k,ℓ
shellingforS (i.e. condition(∗)in[Fu1,p. 101]issatisfied)becausedims ∩s =m+n−2
i,j k,ℓ
ifandonlyifi=kand j6=ℓori6=kand j=ℓ,sot =s ∩ s ∩ s . Hence,
i,j i,j k>i k,j ℓ>j i,ℓ
thecollection{[V(t )]}formsabasisforA∗(X) .
i,j Q (cid:0)T (cid:1) (cid:0)T (cid:1)
SetD :=D ···D ·D ···D ;theemptyproductD =1istheunitinA∗(X) .
(−i,j) −i+1 −1 0 j−1 (−1,0) Q
The generators of M imply that D =0 in A∗(X) if i>m or j >n. Since t is spanned
(−i,j) Q i,j
by the rays r with −i <ℓ< j, it follows that [V(t )]=D . Thus, D for 1 6i6m
ℓ i,j (−i,j) (−i,j)
and 06 j 6n forms a basis for A∗(X) . The degree of a zero-dimensionalsubschemeY of X,
Q
denoteddeg(Y), istherationalnumbersuch that[Y]=deg(Y)D inAm+n−1(X) .
(−m,n) Q
Thefollowingcalculationis thekey toprovingTheorem 2.
4 D.ERMAN,G.G.SMITH,ANDA.VÁRILLY-ALVARADO
k
Lemma 3. For16k6m+n−1,we havek!Dk = (cid:229) k D in A∗(X) .
0 i−1 (−i,k−i+1) Q
i=1
(cid:10) (cid:11)
Proof. In the polynomial ring Q[z], Worpitzky’s identity is zk = (cid:229) k z+i ; see eq. (6.37) in
i i k
[GKP, p. 255] or for a combinatorial proof see [BG, §7]. Rearranging, and homogenizing this
(cid:10) (cid:11)(cid:0) (cid:1)
identitygivestheequationk!zk =(cid:229) k k z+(i−1)y z+(i−2)y ··· z+(i−k)y ,inthe
i=1 i−1
Z-graded polynomialringQ[z,y]withdeg(z)=deg(y)=1. Underthesubstitutionz7→D and
0
(cid:10) (cid:11)(cid:0) (cid:1)(cid:0) (cid:1) (cid:0) (cid:1)
y7→D −D , weobtaintheequation
1 0
k!Dk = (cid:229) k k 1−(i−1) D −(i−1)D ··· 1−(i−k) D −(i−k)D
0 i−1 0 1 0 1
i=1(cid:28) (cid:29)
(cid:16) (cid:17) (cid:16) (cid:17)
(cid:0) (cid:1) (cid:0) (cid:1)
in A∗(X) . To complete the proof, we observe that the ideal L contains the linear relation
Q
D =(1−i)D −iD forall −m6i6n. (cid:3)
i 0 1
Usingthislemma,wecan computethedegreeofcertain completeintersectionsinX.
Proposition 4. Let g ,...,g be homogeneous elements of S such that deg(g )= j for
1 m+n−1 j 0
16 j6m+n−1. If V (g ,...,g )isazero-dimensionalsubschemeofX,thenitsdegree
X 1 m+n−1
(cid:2) (cid:3)
is m+n−1 .
m−1
Pr(cid:10)oof. Ea(cid:11)chhomogeneouspolynomialg definesahypersurfaceinX. ThisCartierdivisorisra-
j
tionallyequivalentto jD becausewehavedeg(g )= j for16 j6m+n−1. Thesubscheme
0 j 0
Z :=V (g ,...,g )has dimensionzero ifand onlyifitisacompleteintersection. Hence,
X 1 m+n−1
(cid:2) (cid:3)
the degree of Z equals the appropriate intersection number, namely the coefficient of D
(−m,n)
in (cid:213) m+n−1 jD ; see Proposition 7.1 in [Fu2]. Since D =0 for i>m or k−i+1>n,
j=1 0 (−i,k−i+1)
Lemma3 yields(cid:213) m+n−1 jD =(m+n−1)!Dm+n−1 = m+n−1 D . (cid:3)
j=1 0 0 m−1 (−m,n),
Proofof Theorem 2. Applying Conjecture 1 for the pai(cid:10)rs of po(cid:11)sitive integers (m,n−1) and
(m−1,n), we see that V (J)∩V (x x ) = ∅. It follows that [V (J)] belongs to the socle
X X −m n X
of A∗(X) and thus V (J) has dimension zero. Since x x is a nonzerodivisoron V (J), we
Q X −m n X
see that deg(I ) equals degV (J); see §2.2. Therefore, applying Proposition 4 completes the
m,n X
proof. (cid:3)
3. SPARSE LAURENT POLYNOMIALS
Inthissection,wecomputethedegreeofsubschemesofX(m,n)correspondingtocertainsparse
Laurent polynomials. Given the recurrence relation that these degrees satisfy, they may be re-
garded as a generalized form of Eulerian numbers. This computation also generates a decom-
positionof m+n−1 as asumofnonnegativeintegers;see(♥).
m−1
Fix a pair of positiveintegers (m,n) and let d be a positiveinteger dividingm+n. Consider
(cid:10) (cid:11)
theclosedsubschemeX ofX correspondingtoLaurent polynomialsoftheform
d
x z−m+x z−m+d+···+x zn−d+x zn.
−m −m+d n−d n
In other words, X is the subscheme of X defined by the monomial ideal generated by the
d
variablesnotbelongingto{x ,x ,...,x ,x }. When d =1,wehaveX =X.
−m −m+d n−d n d
LAURENTPOLYNOMIALSANDEULERIANNUMBERS 5
For16 j6m+n−1,letg beagenericpolynomialinSofdegree j . Thesegenericpoly-
j 0
nomials cut out the subscheme Z :=V (g ,...,g ). Consider Z :=Z∩X . To compute
X 1 m+n−1 d d
(cid:2) (cid:3)
thedegreeofZ , weintroducethefollowingnotation. If06ℓ6d−1, thenwedefine
d
d−1 0 ifgcd(ℓ+1,d)6=1,
:=
(cid:28) ℓ (cid:29)d (1 ifgcd(ℓ+1,d)=1,
k
and weextendthedefinitionof foralltriples(k,ℓ,d)such thatd dividesk+1 via
ℓ d
k (cid:10) (cid:11) k−d k−d
:=(ℓ+1) +(k−ℓ) .
ℓ ℓ ℓ−d
(cid:28) (cid:29)d (cid:28) (cid:29)d (cid:28) (cid:29)d
It followsthat k = k .
ℓ ℓ 1
Proposition 5.(cid:10)T(cid:11)he s(cid:10)ch(cid:11)eme Z has dimension zero and degree m+n−1 when gcd(d,n)= 1;
d m−1 d
otherwisetheschemeZ is empty.
d
(cid:10) (cid:11)
Before proving this proposition, we record a technical lemma. LetW be the vector space of
i
allpolynomialsinS ofdegree i withsupportcontainedin {x ,x ,...,x ,x }. Given
0 −m −m+d n−d n
a subset S ⊆ {d,2d,...,m+n−d}, let D(S) be the subscheme of X defined by the ideal
d
(cid:2) (cid:3)
generated byW foralli∈S.
i
Lemma 6. IfS ⊆{d,2d,...,m+n−d}, thendimD(S)6 m+n−1−|S|.
d
Proof. It suffices to show that D(S) is contained in a finite union of subschemes with dimen-
sion m+n−1−|S|. To a point P=[p : p :...: p ] in the subscheme D(S), we asso-
d −m −m+d n
ciatethesupportsetsE :={i>0| p 6=0}andE :={i>0| p 6=0}. Fromthedefinitionof
+ i − −i
X ,wededucethatE ⊆{m,m−d,...}andE ⊆{n,n−d,...}. ObservethatPliesinthesub-
d + −
spacedefinedbytheidealhx |i∈{−m,−m+d,...,n}\(E ∪E )}iandthatthissubspacehas
i + −
dimension|E |+|E |−2. Hence,itisenoughtoprove|E |+|E |−26 m+n−1−|S|=|S∁|
+ − + − d
whereS∁ :={d,2d,...,m+n−d}\S. Toaccomplishthis,weconsidertheset
P :={i+ j|i∈E , j ∈E , and i+ j6m+n−d}⊆{d,2d,...,m+n−d}.
+ −
To conclude,oneverifies thatP ⊆S∁ and that|E |+|E |−26|P|. (cid:3)
+ −
Sketchof theProofforProposition5. Tobegin,weassumethatgcd(d,n)=1. LetP(W)bethe
productP(W )×P(W )×···×P(W ) and considertheincidencevariety
d 2d m+n−d
U := P,(h ,...,h ) |h (P)=···=h (P)=0 ⊆X ×P(W)
d m+n−d d m+n−d d
with projection(cid:8)m(cid:0)aps p 1: U →Xd an(cid:1)d p 2: U →P(W). We claim tha(cid:9)t dimU 6dimP(W). To
seethis,observethatageneralpointQinX doesnotbelongtothebaselocusofanyW,sothe
d i
fiberp −1(Q)hasdimensiondimP(W)−m+n+1. Onemustalsoconsiderthedimensionsofthe
1 d
various p −1 D(S) , but Lemma 6 shows that none of these preimages has dimension greater
1
than dimP(W). Since Z equals the fiber of p over a general point of P(W), the inequality
d 2
(cid:0) (cid:1)
dimU 6 dimP(W) implies that Z has dimension zero. The appropriate modifications to the
d
proofsofLemma3and Proposition4 showthatthedegreeofZ is m+n−1 .
d m−1 d
(cid:10) (cid:11)
6 D.ERMAN,G.G.SMITH,ANDA.VÁRILLY-ALVARADO
Assumethat e:=gcd(d,n)>1. If m′ :=m/e, n′ :=n/e, and d′ :=d/e, then there is an iso-
morphismX =X(m,n) −→∼= X(m′,n′) =X′ . Underthisidentification,Z isdeterminedbythe
d d d′ d′ d
ideal hg ,g ,...,g i. LetU′ be the incidence variety for the parameters (m′,n′,d′).
d′ 2d′ e(n′+m′)−d′
FromtheproofofLemma6,wededucethatx x isanonzerodivisoronthetopdimensional
−m′ n′
componentsofU′. Hence, the generic polynomialg is also a nonzerodivisoronU′, so the
m′+n′
intersection of the general fibre of p ′: U′ → P(W) with the hypersurface defined by g is
2 m′+n′
empty. Therefore, wehaveZ =∅. (cid:3)
d
To obtain a decomposition for the Eulerian numbers, we stratify the generic complete inter-
sectionZ bysingularitytype. LetX◦ betheopensubschemeofX consistingofallsingularities
d d
of type B(Z/dZ) in X. Each point in Z belongs to X for some d that divides m+n. Setting
d
Z◦ :=Z∩X◦, weobtain
d d
m+n−1 (cid:229)
=deg(Z)= deg(Z◦). (♥)
m−1 d
(cid:28) (cid:29) d|m+n
Moreover,Möbiusinversionand Proposition5 yield
deg(Z◦)=(cid:229) m (c) m+n−1 ,
d c|(m+n)/d m−1
(cid:28) (cid:29)cd
where m is theclassical Möbiusfunction;seeeq. (4.55)and (4.56)in[GKP,p. 136].
The equation (♥) has an elegant combinatorial refinement which we learnt from Alexan-
der Postnikov; cf. [LP, §6]. To sketch this refinement, we observe that the Eulerian number
n also counts the circular permutations of {0,...,n} with k+1 circular ascents. The group
k
Z/(n+1)Z naturally acts on this subset of circular permutations; add 1 modulo n+1 to each
(cid:10) (cid:11)
element. The cardinalities of the orbits then give rise to (♥). More precisely, deg(Z◦) equals
d
the product of (m+n)/d and the number of orbits with cardinality (m+n)/d. For example,
if m = 2 and n = 3, then we have 4 = 11, deg(Z◦) = 1, and deg(Z◦) = 10 = 2·5. On the
1 5 1
other hand, the eleven circular permutations of {0,...,4} with two circular ascents are parti-
(cid:10) (cid:11)
tioned into three Z/5Z-orbits, namely {03241}, {01432,04312,04231,03421,03214},
and {02143,04132,02431,04213,03241}.
4. FURTHER QUESTIONS
4.1. REGULAR SEQUENCE. Theorem2underscoresthesignificanceofConjecture1. Toprove
this conjecture, it would be enough to show that V (Jf˜1K,...,Jf˜m+nK) is the empty set. From
X
thisperspective,theproofofProposition5couldbeviewedasevidencesupportingthisconjec-
ture: for generic elements g of S with degree j , the subscheme V (g ,...,g ) is indeed
j 0 X 1 m+n
empty. On the other hand, Conjecture 1 is false over a field with positive characteristic. For
(cid:2) (cid:3)
instance,if f :=z−1+z∈F [z,z−1],thenwehaveJfiK=0foralli. EvenifConjecture1holds,
2
the F -vector space F [x ,...,x ]/I may fail to have a finite dimension; this happens
p p −m+1 n−1 m,n
when p=2,m=1, andn=2.
LAURENTPOLYNOMIALSANDEULERIANNUMBERS 7
4.2. COMBINATORICS. The positivity and simplicity of many formulae in this article suggest
that we have uncovered only part of the combinatorial structure. To help orient the search for
furtherstructure,weposetwospecific questions:
• Can one find an explicit basis for C[x ,...,x ]/I together with a bijection to
−m+1 n−1 m,n
thepermutationsof[m+n−1]withexactlym−1 ascents?
• Does (cid:229) dim S = m+n−1 hold for all positive m and n? When m=3
C hg1,...,gm+ni j m−1
j>0 0
andn=3,weh(cid:0)ave 5 =6(cid:1)6(cid:2)=(cid:3)1+(cid:10)0+2+(cid:11)3+6+7+9+10+9+7+6+3+2+0+1.
2
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